Concentration Terms & Volumetric Calculations (Molarity, Molality, Normality, Mole fraction, ppm, Strength of solution, Volumetric calculations): NEET MCQs with Answers

15 free questions on Concentration Terms & Volumetric Calculations (Molarity, Molality, Normality, Mole fraction, ppm, Strength of solution, Volumetric calculations) for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

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Question 1

Previous-year question
A 100 mL mixture of NaOH and Na₂SO₄ is completely neutralised by 10 mL of 0.5 M H₂SO₄. What mass of NaOH is present in the 100 mL solution? [NEET 2021 (Paper-2)]
  1. 0.2 g
  2. 0.4 g
  3. 0.6 g
  4. 0.8 g
Show answer
Answer: (B) 0.4 g
Explanation: Only NaOH reacts: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O. Moles of H₂SO₄ = 0.010 L × 0.5 M = 5 × 10⁻³ mol, so moles of NaOH = 2 × 5 × 10⁻³ = 1 × 10⁻² mol. Mass of NaOH = 0.01 × 40 = 0.4 g.

Question 2

Previous-year question
1 g of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution. The mass of sodium hydroxide left unreacted is equal to [NEET 2024]
  1. 750 mg
  2. 250 mg
  3. zero mg
  4. 200 mg
Show answer
Answer: (B) 250 mg
Explanation: NaOH + HCl ⟶ NaCl + H₂O (reacting in a 1 : 1 mole ratio). n(NaOH) = 1/40 = 0.025 mol. n(HCl) = 0.75 × 25/1000 = 0.01875 mol. Comparing (given moles)/(stoichiometric coefficient): NaOH = 0.025, HCl = 0.01875. HCl is smaller, so HCl is the limiting reagent. NaOH consumed = 0.01875 mol; NaOH left = 0.025 − 0.01875 = 0.00625 mol. Mass left = 0.00625 × 40 = 0.25 g = 250 mg.

Question 3

Previous-year question
A 5 molar solution of H₂SO₄ is diluted from 1 litre to a volume of 10 litres. The normality of the resulting solution will be [AIPMT 1991]
  1. 1 N
  2. 0.1 N
  3. 5 N
  4. 0.5 N
Show answer
Answer: (A) 1 N
Explanation: 5 M H₂SO₄ (basicity 2) is 10 N. On dilution equivalents are conserved: N₁V₁ = N₂V₂ → 10 × 1 = N₂ × 10, so N₂ = 1 N (0.5 N would be the molarity, not the normality).

Question 4

Previous-year question
The molarity of liquid HCl, if the density of the solution is 1.17 g/cc, is [AIPMT 2001]
  1. 36.5
  2. 18.25
  3. 32.05
  4. 42.10
Show answer
Answer: (C) 32.05
Explanation: For pure liquid HCl, 1 cc of the liquid contains 1.17 g of HCl. Therefore 1 L (1000 cc) contains 1170 g of HCl. Molar mass of HCl = 1 + 35.5 = 36.5 g/mol. Molarity = (mass of solute × 1000)/(molar mass × volume of solution in mL) = (1.17 × 1000)/(36.5 × 1) = 32.05 M.

Question 5

Previous-year question
Concentrated aqueous sulphuric acid is 98% H₂SO₄ by mass and has a density of 1.80 g mL⁻¹. The volume of this acid required to make one litre of 0.1 M H₂SO₄ solution is [AIPMT 2007]
  1. 16.65 mL
  2. 22.20 mL
  3. 5.55 mL
  4. 11.10 mL
Show answer
Answer: (C) 5.55 mL
Explanation: Molarity of the concentrated acid = (% by mass × density × 10)/molar mass = (98 × 1.80 × 10)/98 = 18.0 M (≈ 18.02 M). Check: 1 L of concentrated acid weighs 1800 g, of which 98% = 1764 g is H₂SO₄ ⇒ 1764/98 = 18 mol. ✓ Moles required in the dilute solution = 0.1 × 1 = 0.1 mol. Using the dilution relation M₁V₁ = M₂V₂: 18.02 × V₁ = 0.1 × 1000 ⇒ V₁ = 100/18.02 = 5.55 mL.

Question 6

Previous-year question
25.3 g of sodium carbonate, Na₂CO₃, is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, the molar concentrations of sodium ion, Na⁺, and carbonate ion, CO₃²⁻, are respectively (molar mass of Na₂CO₃ = 106 g mol⁻¹) [AIPMT 2010]
  1. 0.955 M and 1.910 M
  2. 1.910 M and 0.955 M
  3. 1.90 M and 1.910 M
  4. 0.477 M and 0.477 M
Show answer
Answer: (B) 1.910 M and 0.955 M
Explanation: n(Na₂CO₃) = 25.3/106 = 0.2387 mol in 0.250 L, so [Na₂CO₃] = 0.955 M. Since Na₂CO₃ → 2Na⁺ + CO₃²⁻, [Na⁺] = 2 × 0.955 = 1.910 M and [CO₃²⁻] = 0.955 M (in that order).

Question 7

Previous-year question
6.02 × 10²⁰ molecules of urea are present in 100 mL of its solution. The concentration of the solution is [NEET 2013]
  1. 0.02 M
  2. 0.01 M
  3. 0.001 M
  4. 0.1 M
Show answer
Answer: (B) 0.01 M
Explanation: Moles of urea = (number of molecules)/Nₐ = (6.02 × 10²⁰)/(6.023 × 10²³) = 1 × 10⁻³ mol. Volume of solution = 100 mL = 0.1 L. Molarity = moles of solute / volume of solution in L = (1 × 10⁻³)/0.1 = 1 × 10⁻² = 0.01 M.
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Question 8

Previous-year question
How many grams of concentrated nitric acid should be used to prepare 250 mL of 2.0 M HNO₃? The concentrated acid is 70% HNO₃ by mass. [NEET 2013]
  1. 45.0 g conc. HNO₃
  2. 90.0 g conc. HNO₃
  3. 70.0 g conc. HNO₃
  4. 54.0 g conc. HNO₃
Show answer
Answer: (A) 45.0 g conc. HNO₃
Explanation: Moles of HNO₃ needed = M × V(L) = 2.0 × 250/1000 = 0.5 mol. Molar mass of HNO₃ = 1 + 14 + (3 × 16) = 63 g/mol. Mass of pure HNO₃ required = 0.5 × 63 = 31.5 g. The concentrated acid is only 70% HNO₃ by mass, so mass of concentrated acid = 31.5 × 100/70 = 45.0 g.

Question 9

Previous-year question
The density of a 2 M aqueous solution of NaOH is 1.28 g/cm³. The molality of the solution is [Given that the molar mass of NaOH = 40 g mol⁻¹] [Odisha NEET 2019]
  1. 1.20 m
  2. 1.56 m
  3. 1.67 m
  4. 1.32 m
Show answer
Answer: (C) 1.67 m
Explanation: Take exactly 1 L (1000 cm³) of the solution as a basis. Mass of solution = density × volume = 1.28 × 1000 = 1280 g. A 2 M solution contains 2 mol of NaOH in 1 L ⇒ mass of solute = 2 × 40 = 80 g. Mass of solvent (water) = 1280 − 80 = 1200 g = 1.2 kg. Molality = moles of solute / mass of solvent in kg = 2/1.2 = 1.67 m. Key point: molarity uses the volume of SOLUTION, while molality uses the mass of SOLVENT only.

Question 10

Previous-year question
The mass of potassium dichromate crystals required to oxidize 750 cm3 of 0.6 M Mohr’s salt solution is (K2Cr2O7 = 294 g/mol, Mohr’s salt = 392 g/mol): [AIPMT 2011]
  1. 0.45 g
  2. 22.05 g
  3. 2.2 g
  4. 0.49 g
Show answer
Answer: (B) 22.05 g
Explanation: Mohr’s salt supplies Fe2+: moles = 0.75 × 0.6 = 0.45 mol. Cr2O72– oxidizes 6 Fe2+, so moles dichromate = 0.45/6 = 0.075 mol. Mass = 0.075 × 294 = 22.05 g.

Question 11

Previous-year question
How many grams of H2O2 are required to convert 0.1 mole PbS to 0.1 mole PbSO4? [AIPMT 2002]
  1. 12.8 g
  2. 13.6 g
  3. 16 g
  4. 3.4 g
Show answer
Answer: (B) 13.6 g
Explanation: PbS + 4H2O2 → PbSO4 + 4H2O (S2– → S6+ loses 8 e–, each H2O2 accepts 2 e–). 0.1 mol PbS needs 0.4 mol H2O2. Mass = 0.4 × 34 = 13.6 g.

Question 12

Previous-year question
Which of the following is dependent on temperature? (NEET 2017)
  1. Molarity
  2. Mole fraction
  3. Weight percentage
  4. Molality
Show answer
Answer: (A) Molarity
Explanation: Molarity is moles of solute per litre of solution, and the volume of a solution changes with temperature, so molarity depends on temperature. Mole fraction, weight percentage and molality are based only on masses, so they do not.

Question 13

Two aqueous solutions, one of glucose and one of urea, are both exactly 0.50 M at the same temperature. Which quantity must be the same for the two solutions?
  1. The percentage by mass of the solute
  2. The mass of solute dissolved per litre of solution
  3. The number of solute molecules in 200 g of each solution
  4. The number of solute molecules in 200 mL of each solution
Show answer
Answer: (D) The number of solute molecules in 200 mL of each solution
Explanation: Molarity is moles of solute per litre of solution, so equal molarity fixes the moles of solute in equal volumes. Glucose and urea have different molar masses, so the mass of solute per litre and the mass percentage differ. The two solutions also differ in density, so equal masses of them do not have equal volumes and need not hold equal numbers of molecules.

Question 14

Water is added to 100 mL of a solution of density 1.6 g mL⁻¹ containing 50% acid by mass, until the density of the solution falls to 1.2 g mL⁻¹. Assuming that volumes are additive, what is the mass percentage of acid in the diluted solution?
  1. 16.7%
  2. 33.3%
  3. 22.2%
  4. 25.0%
Show answer
Answer: (C) 22.2%
Explanation: The original solution has a mass of 160 g and holds 80 g of acid. If V mL of water is added, (160 + V)/(100 + V) = 1.2, which gives V = 200 mL. The final mass is 360 g, so the acid is 80/360 × 100 = 22.2% by mass.

Question 15

A water sample contains 62 ppm by mass of nitrate ions, NO₃⁻ (molar mass 62 g mol⁻¹). What are the mass percentage of nitrate and its approximate molality in this dilute solution?
  1. 0.0062% and 1.0 × 10⁻³ m
  2. 0.062% and 1.0 × 10⁻³ m
  3. 0.0062% and 1.0 × 10⁻² m
  4. 0.00062% and 1.0 × 10⁻⁴ m
Show answer
Answer: (A) 0.0062% and 1.0 × 10⁻³ m
Explanation: 62 ppm means 62 g of nitrate in 10⁶ g of solution, so the mass percentage is 62 × 10⁻⁶ × 100 = 0.0062%. This is also 62 mg, or 1.0 × 10⁻³ mol, per kilogram of solution. The solution is so dilute that 1 kg of it is practically 1 kg of water, so the molality is about 1.0 × 10⁻³ m.

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